Circles
Why This Matters
Watch a bicycle wheel roll along the road. At every instant it touches the ground at exactly one point, and the road behaves like a line that just grazes the circular wheel. Look at the rope running over a pulley at a well: each side of the rope leaves the pulley along a line that touches the wheel at a single point. These touching lines have a special name — tangents — and they obey two beautiful rules that turn up again and again in geometry, physics and design.
In Class 9 you learned what a circle is: all the points at a fixed distance (the radius) from a fixed centre. Now we ask a sharper question: when a straight line and a circle share the same plane, how can they meet? The line can miss the circle entirely, slice straight through it at two points, or just kiss it at one point. That last case — the tangent — is the whole story of this chapter.
The payoff is two facts you can lean on forever: a tangent is always at a perfect right angle to the radius drawn to the touch point, and the two tangents you draw from a single outside point are exactly equal in length. We won’t just state these — we’ll prove them, so you know they are true and not merely “what the diagram looks like”.
The Big Idea
When a line and a circle live in the same plane, they meet in one of only three ways: the line misses the circle (no common point — a non-intersecting line), cuts it at two points (a secant), or touches it at exactly one point (a tangent). A tangent is the limiting case of a secant whose two crossing points have slid together into one. At that single touch point, the radius and the tangent are always perpendicular, and from any point outside the circle you can draw exactly two tangents — and they are equal in length.
Let’s Break It Down
A line and a circle: three possibilities
Take a circle and a straight line PQ in the same plane and slide the line around. Only three situations are ever possible:
- No common point — the line stays clear of the circle. It is a non-intersecting line.
- Two common points — the line crosses into the circle and out again, cutting it at two points A and B. This line is a secant.
- Exactly one common point — the line just touches the circle. This line is a tangent, and the single shared point is the point of contact.
There is no fourth possibility. The word tangent comes from the Latin tangere, “to touch”.
A tangent is really a secant pushed to its limit. Imagine sliding a secant outward, keeping it parallel to itself: the two crossing points creep closer and closer until they merge into one. At that instant the secant has become a tangent.
A tangent to a circle is a special case of a secant, taken when the two end points of the chord it cuts off come together into a single point.
A line meets a circle at two distinct points. What is this line called, and is it a tangent?
How many tangents pass through a given point?
Where the point sits relative to the circle decides everything:
- Point inside the circle — every line through it cuts the circle at two points, so no tangent can pass through a point inside the circle.
- Point on the circle — there is exactly one tangent at that point.
- Point outside the circle — you can draw exactly two tangents to the circle from that point.
| Where the point is | Number of tangents |
|---|---|
| Inside the circle | none (0) |
| On the circle | exactly one (1) |
| Outside the circle | exactly two (2) |
The length of the piece of tangent from an external point P up to the point of contact is called the length of the tangent from P. We’ll prove a lovely fact about it shortly.
Theorem 1 — The tangent is perpendicular to the radius at the point of contact
Statement. The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Given. A circle with centre O and a tangent XY touching the circle at the point P.
To prove. OP ⊥ XY.
Proof. Take any point Q on the line XY other than P, and join OQ.
Now Q cannot lie on the circle — if it did, the line XY would meet the circle at two points (P and Q) and would be a secant, not a tangent. And Q cannot lie inside the circle for the same reason (a line through an inside point cuts the circle twice). So Q lies outside the circle, which means OQ is longer than the radius:
OQ > OP.
This is true for every point Q on XY except P itself. So among all the points of the line XY, the point P is the one closest to O — that is, OP is the shortest distance from O to the line XY.
But the shortest distance from a point to a line is always the perpendicular distance. Therefore OP must be perpendicular to XY.
OP ⊥ XY. ∎
Two useful spin-offs:
- At any point on a circle there is one and only one tangent.
- The line through the radius at the point of contact is sometimes called the normal to the circle at that point.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Find the length PQ.
- P is the point of contact, so OP is a radius: OP = 5 cm. By Theorem 1, the tangent PQ is perpendicular to the radius OP, so ∠OPQ = 90°.
- That makes △OPQ a right triangle with the right angle at P. Here OQ is the hypotenuse (the side opposite the right angle), and OP, PQ are the two legs.
- By the Pythagoras theorem: OQ² = OP² + PQ², so PQ² = OQ² − OP² = 12² − 5² = 144 − 25 = 119.
- Therefore PQ = √119 cm ≈ 10.9 cm. (Notice 5, 12 are not a Pythagorean pair here because 12 is the hypotenuse, not a leg — so the answer is the slightly awkward √119, not a whole number.)
Theorem 2 — The two tangents from an external point are equal
Statement. The lengths of the tangents drawn from an external point to a circle are equal.
Given. A circle with centre O, a point P lying outside the circle, and two tangents PA and PB drawn from P, touching the circle at A and B respectively.
To prove. PA = PB.
Proof. Join OA, OB and OP.
Because PA is a tangent and OA is the radius to the point of contact A, Theorem 1 gives ∠OAP = 90°. Likewise PB is a tangent and OB its radius, so ∠OBP = 90°. So △OAP and △OBP are right-angled triangles, each with its right angle at the point of contact.
Now compare the right triangles △OAP and △OBP:
- OA = OB — both are radii of the same circle.
- OP = OP — this side is common to both triangles (it is the hypotenuse of each).
- ∠OAP = ∠OBP = 90°.
So by the RHS congruence rule (Right angle – Hypotenuse – Side), △OAP ≅ △OBP.
Since corresponding parts of congruent triangles are equal (CPCT):
PA = PB. ∎
A quick alternative using Pythagoras. In each right triangle, PA² = OP² − OA² and PB² = OP² − OB². But OA = OB (radii), so PA² = PB², giving PA = PB.
A bonus fact. From the congruence we also get ∠OPA = ∠OPB. So OP bisects the angle ∠APB between the two tangents — the centre always lies on the bisector of the angle between the two tangents.
A quadrilateral ABCD is drawn so that all four of its sides touch a circle (the circle is inscribed in it). Prove that AB + CD = AD + BC.
- Let the circle touch sides AB, BC, CD, DA at the points P, Q, R, S respectively. From each vertex, two tangents to the circle are drawn — the two sides meeting there.
- By Theorem 2 (equal tangents from an external point), the two tangent lengths from each vertex are equal: from A, AP = AS; from B, BP = BQ; from C, CR = CQ; from D, DR = DS.
- Add the left sides and the right sides of those four equalities: (AP + BP) + (CR + DR) = (AS + BQ) + (CQ + DS). The left group is AB + CD; regroup the right group as (AS + DS) + (BQ + CQ) = AD + BC.
- Therefore AB + CD = AD + BC — in any quadrilateral that circumscribes a circle, the two sums of opposite sides are equal. ∎
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2 ∠OPQ.
- Let ∠PTQ = θ. By Theorem 2, TP = TQ, so △TPQ is isosceles with TP = TQ.
- The base angles of an isosceles triangle are equal, and the three angles sum to 180°: ∠TPQ = ∠TQP = ½(180° − θ) = 90° − ½θ.
- By Theorem 1, the radius OP is perpendicular to the tangent TP, so ∠OPT = 90°.
- Now ∠OPQ = ∠OPT − ∠TPQ = 90° − (90° − ½θ) = ½θ = ½∠PTQ. Multiplying by 2 gives ∠PTQ = 2 ∠OPQ. ∎
Common Mistakes
A tangent and the radius at the point of contact meet at some general angle that depends on the circle.
Different-sized circles and different tangent lines look so varied that it seems the angle should change from picture to picture.
It is ALWAYS exactly 90°, for every circle and every tangent. Theorem 1 proves the radius to the contact point is the shortest distance from the centre to the tangent line, and the shortest distance is always perpendicular.
In the radius-tangent right triangle, OQ² = OP² + PQ² means OQ is just √(OP² + PQ²), so PQ is a leg you can find by adding.
Students remember 'Pythagoras = add the squares' and apply it without checking which side is the hypotenuse.
The RIGHT ANGLE is at the contact point P (radius ⊥ tangent), so OQ — the line to the centre — is the HYPOTENUSE. You SUBTRACT: PQ² = OQ² − OP². Adding gives the wrong, larger answer.
You can draw two tangents to a circle from any point, inside or outside.
The 'two equal tangents' result is so memorable that it feels like it should hold everywhere.
Two tangents come only from a point OUTSIDE the circle. From a point ON the circle there is exactly one tangent; from a point INSIDE there are none (every line through an inside point cuts the circle twice).
The two tangents from an external point only look equal in neat textbook figures; in a lopsided drawing they would differ.
A hand-drawn or skewed figure can make PA look longer than PB, so equality feels like an artifact of careful drawing.
PA = PB is a proven theorem (RHS congruence of △OAP and △OBP), true for EVERY external point and EVERY circle, no matter how the figure is drawn. The centre even lies on the bisector of the angle between the tangents.
Quick Check
A straight line touches a circle at exactly one point. What is the line called?
The tangent at a point P of a circle with centre O makes what angle with the radius OP?
From a point Q the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. What is the radius of the circle?
Tangents PA and PB are drawn from an external point P to a circle. Which statement is always true?
Practice Problems
Easy
The length of a tangent from a point A at distance 5 cm from the centre of a circle is 4 cm. Find the radius of the circle.
Let the tangent touch the circle at B, with centre O. The radius OB is perpendicular to the tangent AB, so △OBA is right-angled at B.
Here OA = 5 cm is the hypotenuse and AB = 4 cm is one leg. By Pythagoras:
OB² = OA² − AB² = 5² − 4² = 25 − 16 = 9.
So the radius OB = 3 cm.
How many tangents can be drawn to a circle from (i) a point inside it, (ii) a point on it, (iii) a point outside it?
(i) From a point inside the circle: 0 tangents — every line through an interior point cuts the circle at two points.
(ii) From a point on the circle: exactly 1 tangent.
(iii) From a point outside the circle: exactly 2 tangents (and they are equal in length).
Medium
Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Let O be the common centre. The chord AB of the larger circle touches the smaller circle at a point P, so OP is a radius of the smaller circle and is perpendicular to AB (tangent ⊥ radius): OP = 3 cm.
Because OP ⊥ AB and OP is drawn from the centre, OP bisects the chord (the perpendicular from the centre bisects the chord). So P is the midpoint of AB, and AP = PB.
In right triangle OPA, OA = 5 cm (radius of the larger circle) is the hypotenuse:
AP² = OA² − OP² = 5² − 3² = 25 − 9 = 16, so AP = 4 cm.
The full chord AB = 2 × AP = 2 × 4 = 8 cm.
Prove that the tangents drawn at the two ends of a diameter of a circle are parallel.
Let AB be a diameter of a circle with centre O. Draw the tangent at A and the tangent at B.
By Theorem 1, the tangent at A is perpendicular to the radius OA, and the tangent at B is perpendicular to the radius OB. Since AB is a diameter, OA and OB lie along the same straight line AB.
So both tangents are perpendicular to the same line AB. Two lines that are each perpendicular to the same line are parallel to each other.
Therefore the tangents at the ends of a diameter are parallel. ∎
Challenge
PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length TP.
Join OT (O is the centre). Since TP = TQ (equal tangents), △TPQ is isosceles, and TO is the bisector of ∠PTQ. So OT ⊥ PQ and OT bisects PQ at the point R, giving PR = RQ = 4 cm.
In right triangle OPR (right angle at R), OP = 5 cm is the hypotenuse: OR² = OP² − PR² = 5² − 4² = 25 − 16 = 9, so OR = 3 cm.
Now use angles. In right △OPT the radius OP ⊥ tangent TP, so ∠OPT = 90°, meaning ∠TPR + ∠RPO = 90°. In right △TPR (right angle at R), ∠TPR + ∠PTR = 90°. Comparing these, ∠RPO = ∠PTR.
So right triangle TRP is similar to right triangle PRO (AA similarity). Matching corresponding sides:
TP / PO = RP / RO, i.e. TP / 5 = 4 / 3, so TP = 20/3 cm ≈ 6.67 cm.
(Check by Pythagoras with TP = x, TR = y: x² = y² + 16 from △PRT, and x² + 25 = (y + 3)² from △OPT. Subtracting gives y = 16/3, then x² = (16/3)² + 16 = 400/9, so x = 20/3 cm. ✓)
Summary
You should now be able to explain:
- A line in the plane of a circle either misses it, cuts it at two points (a secant), or touches it at exactly one point (a tangent); the touch point is the point of contact.
- A tangent is the limiting case of a secant whose two crossing points have merged into one.
- Tangents through a point: none from inside, exactly one from a point on the circle, exactly two from a point outside.
- Theorem 1: the tangent at any point is perpendicular to the radius through the point of contact (proved because the radius is the shortest distance from the centre to the tangent line).
- Theorem 2: the two tangents from an external point are equal in length (proved by RHS congruence of the two right triangles), and the centre lies on the bisector of the angle between them.
- In any right triangle made by a radius and a tangent, the line to the centre is the hypotenuse — so you subtract the squares to find a tangent length.
What’s Next
You now know how lines touch circles. Next, in Areas Related to Circles, you’ll measure the circle itself — the area and circumference, then the length of an arc, and the areas of sectors and segments (the pizza-slice and the bow-shaped pieces). Those formulas, combined with the tangent facts from this chapter, let you find areas of all sorts of curved figures you meet in design and real life.